Calculus Examples

arcsin (x) + arcsin (y) = \frac{π}{2}

$\arcsin (x) + \arcsin (y) = \frac{π}{2}$ ,

(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})

$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$

Step 1

Subtract

arcsin (y)

$\arcsin (y)$ from both sides of the equation.

arcsin (x) = \frac{π}{2} - arcsin (y)

$\arcsin (x) = \frac{π}{2} - \arcsin (y)$

Step 2

Rewrite the equation as

\frac{π}{2} - arcsin (y) = arcsin (x)

$\frac{π}{2} - \arcsin (y) = \arcsin (x)$ .

\frac{π}{2} - arcsin (y) = arcsin (x)

$\frac{π}{2} - \arcsin (y) = \arcsin (x)$

Step 3

Subtract

\frac{π}{2}

$\frac{π}{2}$ from both sides of the equation.

- arcsin (y) = arcsin (x) - \frac{π}{2}

$- \arcsin (y) = \arcsin (x) - \frac{π}{2}$

Step 4

Divide each term in

- arcsin (y) = arcsin (x) - \frac{π}{2}

$- \arcsin (y) = \arcsin (x) - \frac{π}{2}$ by

- 1

$- 1$ and simplify.

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Step 4.1

Divide each term in

- arcsin (y) = arcsin (x) - \frac{π}{2}

$- \arcsin (y) = \arcsin (x) - \frac{π}{2}$ by

- 1

$- 1$ .

\frac{- arcsin (y)}{- 1} = \frac{arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\frac{- \arcsin (y)}{- 1} = \frac{\arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

\frac{arcsin (y)}{1} = \frac{arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\frac{\arcsin (y)}{1} = \frac{\arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

Step 4.2.2

Divide

arcsin (y)

$\arcsin (y)$ by

1

$1$ .

arcsin (y) = \frac{arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\arcsin (y) = \frac{\arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

arcsin (y) = \frac{arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\arcsin (y) = \frac{\arcsin (x)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Move the negative one from the denominator of

\frac{arcsin (x)}{- 1}

$\frac{\arcsin (x)}{- 1}$ .

arcsin (y) = - 1 \cdot arcsin (x) + \frac{- \frac{π}{2}}{- 1}

$\arcsin (y) = - 1 \cdot \arcsin (x) + \frac{- \frac{π}{2}}{- 1}$

Step 4.3.1.2

Rewrite

- 1 \cdot arcsin (x)

$- 1 \cdot \arcsin (x)$ as

- arcsin (x)

$- \arcsin (x)$ .

arcsin (y) = - arcsin (x) + \frac{- \frac{π}{2}}{- 1}

$\arcsin (y) = - \arcsin (x) + \frac{- \frac{π}{2}}{- 1}$

Step 4.3.1.3

Dividing two negative values results in a positive value.

arcsin (y) = - arcsin (x) + \frac{\frac{π}{2}}{1}

$\arcsin (y) = - \arcsin (x) + \frac{\frac{π}{2}}{1}$

Step 4.3.1.4

Divide

\frac{π}{2}

$\frac{π}{2}$ by

1

$1$ .

arcsin (y) = - arcsin (x) + \frac{π}{2}

$\arcsin (y) = - \arcsin (x) + \frac{π}{2}$

arcsin (y) = - arcsin (x) + \frac{π}{2}

$\arcsin (y) = - \arcsin (x) + \frac{π}{2}$

arcsin (y) = - arcsin (x) + \frac{π}{2}

$\arcsin (y) = - \arcsin (x) + \frac{π}{2}$

arcsin (y) = - arcsin (x) + \frac{π}{2}

$\arcsin (y) = - \arcsin (x) + \frac{π}{2}$

Step 5

Take the inverse arcsine of both sides of the equation to extract

x

$x$ from inside the arcsine.

y = sin (- arcsin (x) + \frac{π}{2})

$y = \sin (- \arcsin (x) + \frac{π}{2})$

Step 6

Rewrite the equation as

sin (- arcsin (x) + \frac{π}{2}) = y

$\sin (- \arcsin (x) + \frac{π}{2}) = y$ .

sin (- arcsin (x) + \frac{π}{2}) = y

$\sin (- \arcsin (x) + \frac{π}{2}) = y$

Step 7

Take the inverse sine of both sides of the equation to extract

arcsin (x)

$\arcsin (x)$ from inside the sine.

- arcsin (x) + \frac{π}{2} = arcsin (y)

$- \arcsin (x) + \frac{π}{2} = \arcsin (y)$

Step 8

Subtract

\frac{π}{2}

$\frac{π}{2}$ from both sides of the equation.

- arcsin (x) = arcsin (y) - \frac{π}{2}

$- \arcsin (x) = \arcsin (y) - \frac{π}{2}$

Step 9

Divide each term in

- arcsin (x) = arcsin (y) - \frac{π}{2}

$- \arcsin (x) = \arcsin (y) - \frac{π}{2}$ by

- 1

$- 1$ and simplify.

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Step 9.1

Divide each term in

- arcsin (x) = arcsin (y) - \frac{π}{2}

$- \arcsin (x) = \arcsin (y) - \frac{π}{2}$ by

- 1

$- 1$ .

\frac{- arcsin (x)}{- 1} = \frac{arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\frac{- \arcsin (x)}{- 1} = \frac{\arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

Step 9.2

Simplify the left side.

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Step 9.2.1

Dividing two negative values results in a positive value.

\frac{arcsin (x)}{1} = \frac{arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\frac{\arcsin (x)}{1} = \frac{\arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

Step 9.2.2

Divide

arcsin (x)

$\arcsin (x)$ by

1

$1$ .

arcsin (x) = \frac{arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\arcsin (x) = \frac{\arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

arcsin (x) = \frac{arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}

$\arcsin (x) = \frac{\arcsin (y)}{- 1} + \frac{- \frac{π}{2}}{- 1}$

Step 9.3

Simplify the right side.

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Step 9.3.1

Simplify each term.

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Step 9.3.1.1

Move the negative one from the denominator of

\frac{arcsin (y)}{- 1}

$\frac{\arcsin (y)}{- 1}$ .

arcsin (x) = - 1 \cdot arcsin (y) + \frac{- \frac{π}{2}}{- 1}

$\arcsin (x) = - 1 \cdot \arcsin (y) + \frac{- \frac{π}{2}}{- 1}$

Step 9.3.1.2

Rewrite

- 1 \cdot arcsin (y)

$- 1 \cdot \arcsin (y)$ as

- arcsin (y)

$- \arcsin (y)$ .

arcsin (x) = - arcsin (y) + \frac{- \frac{π}{2}}{- 1}

$\arcsin (x) = - \arcsin (y) + \frac{- \frac{π}{2}}{- 1}$

Step 9.3.1.3

Dividing two negative values results in a positive value.

$\arcsin (x) = - \arcsin (y) + \frac{\frac{π}{2}}{1}$

Step 9.3.1.4

Divide

$\frac{π}{2}$ by

$1$ .

$\arcsin (x) = - \arcsin (y) + \frac{π}{2}$

Step 10

Take the inverse arcsine of both sides of the equation to extract

$x$ from inside the arcsine.

$x = \sin (- \arcsin (y) + \frac{π}{2})$

Step 11

The equation can not be solved. The given interval accounts for only one variable, but

$3$ are present in the equation

$x = \sin (- \arcsin (y) + \frac{π}{2})$ .

No solution